Question
Question: If \[x\] is a negative integer,then find the solution set of the inequality \[ \dfrac{2}{3} + \dfr...
If x is a negative integer,then find the solution set of the inequality
32+31(x+1)>0
Solution
First find the solution set for the given inequality,and by using the given data in the question subtract the non-negative integer part from the solution set,then you can able to find the solution set of the given problem easily.
Complete step by step answer:
Step 1: Firstly,we need to carefully look at the given data in the problem,so that we are able to reduce our errors while we are solving a problem. Let's consider the given data in the question,they had clearly mentioned that x is a negative integer. Now we have to consider the data given in the hint, firstly we need to find the solution set of the inequality by solving them,then we need to subtract the non-negative part from it.
Step 2:
On further proceeding,according to the given hint. Now we are finding the solution for given inequality i.e, 32+31(x+1)>0
On solving for we will get,
32+31(x+1)>0
Opening the brackets, we get,
⇒32+3x+31>0
Adding the like terms, we get,
⇒3x+1>0
Shifting the constant terms to right side, we get,
⇒3x>−1
Multiplying both sides by 3, we get,
⇒x>−3
This shows that the solution set of the given problem is x∈[−3,∞).
Step 3: According to the given data of the question x is a non-negative integer
By considering the given condition in the question we need to subtract the non-negative integral part from the solution set of the inequality i.e,subtract (0,∞)from the solution set of inequality. This gives the solution of the given problem (−3,∞)−(0,∞)=(−3,0).
Therefore,finally we got the solution set of the equation i.e, x∈(−3,0).
Note: When both sides of an inequality are multiplied with a negative number, then the sign of inequality changes. Students must remember this to obtain the correct answer.When we multiply both sides of an inequality by a positive number, then the sign of inequality remains unchanged. Open intervals means that the end points are not included.